For what values of c will the quadratic equation y=x 2 +6x+ c have no real zeroes? Set up and solve an inequality for this problem. ...thanks for the help

c<9

For what values of c will the quadratic equation y=x2+6x+c have no real zeroes? Set up and solve an inequality for this problem. ...thanks for the help

$\begin{array}{rcl} \boxed{~ \begin{array}{rcl} ax^2+bx+c&=&0\\ x &=& {-b \pm \sqrt{b^2-4ac} \over 2a} \\ \text{discriminant D } &=& b^2-4ac\\ \text{no real zeroes } D < 0 && b^2-4ac < 0 \end{array} ~} \end{array}$

$\begin{array}{rcll} y=x^2+6x+c &=& 0 \qquad a=1 \qquad b=6 \\\\ \boxed{~ \begin{array}{rcl} b^2-4ac &<& 0 \\ \end{array} ~}\\ (-6)^2-4c &<& 0 \\ 36-4c &<& 0 \qquad & | \qquad + 4c \\ 36 &<& 4c \qquad & | \qquad :4 \\ 9 &<& c \\ c &>& 9 \\ \end{array}$

no real zeroes if c > 9